【LeetCode】 143. Reorder List

最新推荐文章于 2022-10-08 18:49:12 发布

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本文介绍了一种链表重组算法，该算法将一个单链表L0→L1→…→Ln-1→Ln重新排列为L0→Ln→L1→Ln-1→L2→Ln-2→…。通过在不改变节点值的情况下完成这一操作，实现了原地重构。文章还提供了完整的Java实现代码。

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Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if (head == null || head.next == null) {
            return;
        }
        ListNode p1 = head, p2 = head;
        while (p2.next != null && p2.next.next != null) {
            p1 = p1.next;
            p2 = p2.next.next;
        }
        ListNode prevMid = p1, curr = p1.next;
        while (curr.next != null) {
            ListNode temp = curr.next;
            curr.next = temp.next;
            temp.next = prevMid.next;
            prevMid.next = temp;
        }
        p1 = head;
        p2 = prevMid.next;
        while (p1 != prevMid) {
            prevMid.next = p2.next;
            p2.next = p1.next;
            p1.next = p2;
            p1 = p2.next;
            p2 = prevMid.next;
        }
    }
}